IIT JEE , AIEEE Forum - Physics , Mechanics

fwks_phoenix

vertical motion under gravity
Particle projected from th etop of a cliff

there r two cases:
when particle is projected upwards ; vertical distance from point O at t = h
then
1. if it is above the point O h=ut-1/2gt2
2.if it is below O -h=ut-1/2gt2

can u explain the second case for me ?? where did the "-" sign cum from if it's throwing upwards ??

n how do u derive the equation
t=u/g+root(u2/g2+2h/g)

using 1/2gt2-ut=h ? (it should b h=ut-1/2gt2 but this is the eqn. given in FIITJEE..plz tell me how this eqn is correct too)

Help me out soon plz..

Anu.

Asmita

suppose if u r on a hill even if u project a ball in d upward direction isnt it possible tht i may go down i.e. below d level of d observer at sum t......... thn accordin 2 d sign convention........ it'll b -h......... thts y its

-h=ut-gt^2 /2 is tht ok???

Asmita

v=u+gt

also v^2=u^2+2gh

so u^2+2ugt+g^2t^2=u^2+2gh

g^2t^2+2ugt-2gh=0

so by using quadratic formula

t= -2ug+tr(4u^2g^2+8g^3t)/2g^2 (since t is always +ve)

=-u/g + rt(u^2/g^2 +2h/g)

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